y = cos(2t), y = sin(2t);

t is greater than or equal to -pie and less than or equal to pie.

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- Jun 24th 2008, 08:06 PMcityismineHow do i convert this into a rectangular equation
y = cos(2t), y = sin(2t);

t is greater than or equal to -pie and less than or equal to pie. - Jun 24th 2008, 09:21 PMmr fantastic
- Jun 25th 2008, 08:34 AMMathstud28
Just so you know, given

$\displaystyle x=a\cos(n\theta)$

and $\displaystyle y=a\sin(n\theta)$

You will always get a circle of radius a

$\displaystyle x^2+y^2=a^2\cos^2(n\theta)+a^2\sin^2(n\theta)=a^2$

and if you have

$\displaystyle x=a\cos(n\theta)$

and

$\displaystyle y=b\sin(n\theta)$

You will get an ellipse

$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{a^2\cos^2(n\ theta)}{a^2}+\frac{b^2\sin^2(n\theta)}{b^2}=1$ - Jun 25th 2008, 09:10 AMgalactusQuote:

**-pie**and less than or equal to**pie**.

- Jun 25th 2008, 06:30 PMcityismine
The answer in the back of the book is: y=(1/2)x^2-1

Is the answer in the book wrong? - Jun 25th 2008, 06:37 PMMathstud28
- Jun 25th 2008, 07:09 PMmr fantastic
If the parametric equations were actually $\displaystyle x = \cos t$ and $\displaystyle y = \cos (2t)$ then the cartesian equation would be $\displaystyle y = 2x^2 - 1$ ........

If the parametric equations were actually $\displaystyle x = 2 \sin t$ and $\displaystyle y = \cos (2t)$ then the cartesian equation would be $\displaystyle y = 1 - \frac{x^2}{2}$ ........

Are the equations you posted correct? - Jun 25th 2008, 08:29 PMcityismine
Yes, I posted them correctly, I just double checked. I guess it's a typo in the book.